Question

$If{\int }_2^e[\frac{1}{logx}-\frac{1}{(logx{)}^2}]dx=\alpha +\frac{\beta }{log2},then$

• A. $\alpha =e,\beta =-2$
• B. $\alpha =e,\beta =2$
• C. $\alpha =-e,\beta =2$
• D. $\alpha =-e,\beta =-2$

Answer 1

The correct option is A $\alpha =e,\beta =-2$

${\int }_2^e[\frac{1}{logx}-\frac{1}{(logx{)}^2}]dx=\alpha +\frac{\beta }{log2}L.H.S.={\int }_2^e[\frac{1}{logx}-\frac{1}{(logx{)}^2}]dx={\int }_2^e\frac{1}{logx}dx-{\int }_2^e\frac{1}{(logx{)}^2}dx=\left[\right(\frac{x}{logx}{)}_2^e-{\int }_2^e\{-\frac{1}{x(logx{)}^2}\}xdx]{\int }_2^e\frac{1}{x(logx{)}^2}dx=|\frac{x}{logx}{|}_2^e=e-\frac{2}{log2}$
Comparing it with the given value, we get $\alpha =e,\beta =-2$